Let the twodigit number be of the form of xy, where x is the ten's digit and y is the unit's digit. The value of this number is (10x + y). The number formed on reversing its digits is of the form of yx. Its value is (10y + x). The difference between these numbers = 9x − y We need this to be a perfect square. ∴ 9x − y = 0, 1, 4, 9, 16, 25, 36, 49, 64 or 81
The perfect squares greater than 81 are ruled out because x − y cannot be greater than 9. We can also rule out 1, 4, 16, 25, 49 and 64 since x − y must be an integer. So, there are only 4 numbers 0, 9, 36, 81 remaining.
The number of 2 digit numbers for which 9x − y = 0, i.e. x = y, comes out to be 9. These numbers are 11, 22, ..., 99.
The number of 2 digit numbers for which 9x − y = 9, i.e. x − y = 1, is calculated as follows: For every value of x from 1 to 8, there will be two values of y such that x − y = 1 For example, for x = 1, we have 10 and 12. But for x = 9 we have only 98. ∴ The number of 2 digit numbers for which x − y = 1 is (twice the number of allowed values of x) − 1 = 2 × 9 − 1 = 17
The number of 2 digit numbers for which 9x − y = 36, i.e. x − y = 4 is calculated as follows: For x = 1 to 3 and 6 to 9, there will only be one value of y for which x − y = 4. For x = 4 and 5, there will be two values of y each (giving the numbers 40, 48, 51 and 59). ∴ The total number of numbers for which x − y = 4 is 7 + 2 + 2 = 11
There is only one 2 digit number for which 9x − y = 81 and it is 90.
∴ The total number of 2 digit numbers satisfying our criteria is (9 + 17 + 11 + 1) = 38
Hence, option 4.
